In many investigations, it becomes necessary to consider the functions of arcs greater than 360°; but since the addition of 360° any number of times to the arc a, will give an arc terminating in the extremity of a, it is obvious that the arc resulting from such addition will have the same functions as the arc a. And hence it follows that the functions of arcs, however great, may be expressed in terms of the functions of arcs less than 90°. SECTION II. PLANE TRIGONOMETRY, PRACTICALLY APPLIED. In the preceding section, the theory of Trigonometry has been quite fully developed, and the student should now be prepared for its various applications, were he acquainted with logarithms. But logarithms are no part of Trigonometry, and serve only to facilitate the numerical operations. Trigonometrical computations can be made without logarithms, and were so made long before the theory of logarithms was understood. For this reason, we proceed at once to the solution of the following triangles. 1. The hypotenuse of a right-angled triangle is 21, and the base is 17; required the perpendicular and the acute angles. We must now turn to Table II, and find in the last two columns the cosine nearest to .80952, and the corresponding degrees and minutes will be the value of the angle C. On page 57, of Tables, near the bottom of the page, and in the column with cosine at the top, we find .80953, which corresponds to 35° 56' for the angle C. The angle B is, therefore, 54° 3'. This Table is so arranged, that the sum of the degrees at the top and bottom of the page, added to the sum of the minutes which are found on the same horizontal line in the two side columns of the page, is 90° Thus, in finding the angle C, the number .80953 was found in the column with cosine at the head. We therefore took the degrees from the head of the page, and the minutes were taken from the left hand column, counting downwards. For the side AB, we have the proportion or, that is, CF: FD :: CA : AB; cos. C sin. C :: 17: AB; .80953: .58708 :: 17 : AB. From which we find AB = .58708 × 17.80953; If we had formed a table of natural tangents, as well as of natu ral sines, AB could have been found by the following proportion⚫ The perpendicular AB may also be found by the proportion or, CD: DF:: CB: AB; 1 : sin. C :: 21 : AB; whence, AB = 21 sin. C = 21 × .58708 = 12.32868. 2. The two sides of a right-angled triangle are 150 and 125; required the hypotenuse and the acute angles. We may employ the same figure as in the preceding problem. Then, from the similar triangles, CFD and CAB, we get CF: FD:: CA: AB; D FE B that is, which gives hence, cos. C sin. C: 150: 125 :: 6 : 5, 36 sin. C = 25 cos. C. Adding member to member, 36 cos.2C = 36 cos.' C. we have 36 (sin.C+cos. C) =61 cos. C. But sin.2C+cos.2 C = 1, (Eq. (1) Trigonometry); To find the angle of which this is the cosine, we turn to page 60 of tables, and looking in the column having cosine at the head, we see that .76822 falls between .76828, which has 48′ opposite to it in the left hand column, and .76810, which has 49' opposite to it in the same column. Now, the cosines of arcs less than 90° decrease when the arcs increase, and the converse; and while the increase of the arc is confined within the limits of 1', the increase of the arc will be sensibly proportional to the decrease of the cosine. B which gives The angle C is, therefore, equal to 39° 48′ 20′′, and the angle = 90° 39° 48′ 20′′ = 50° 11′ 40′′. 3. The base of a right-angled triangle is 150, and the a gle opposite the base is 50° 11′ 40′′; required "the hypotenuse and the perpendicular. 4. Two sides, the one 30 and the other 35, and the included angle 20°, of a triangle, are given, to find the other two angles and the third side. = Let BAC be the triangle, in which BC 35, BA = 30, and the angle B 20°. From A, the extremity of the shorter side, let fall on BC the perpendicular AD, thus dividing the triangle B into the two right-angled triangles BAD and CAD. Then, from the triangle BAD, we have 1st, or, 2d, or, sin. D sin. B :: BA : AD; 1 sin. 20° :: 30 : AD = 30 sin. 20° 1 : cos. B :: BA : BD; 1: cos. 20° :: 30 : BD = 30 cos. 20°. In the table of natural sines, we find sin. 20° = Jos. 20°.93969; hence, AD = 30 × .34202 BD = 30 x .93969 = 28.19070, and therefore BD 6.8093. From the triangle CAD, we have 1st, ACAD2 + DC2 2d, .34202, and the 10.26060, and DC = BC = ✔(10.26)2 + (6.8+)' 12 267 |